[PYTHON] import sympy as sp

Define the variable theta

theta = sp.symbols('theta')

Define the polar curve r

r = 2 + sp.cos(theta)

Part (1): Compute the area a

Area = (1/2) * integral(r^2, theta, 0, 2*pi)

area_expr = (1/2) * r**2 a = sp.integrate(area_expr, (theta, 0, 2*sp.pi)) print(f"Area a: {a}")

Part (2): Compute the slope at theta = pi/4

x = r * cos(theta), y = r * sin(theta)

x = r * sp.cos(theta) y = r * sp.sin(theta)

Derivatives with respect to theta

dx_dtheta = sp.diff(x, theta) dy_dtheta = sp.diff(y, theta)

Slope dy/dx = (dy/dtheta) / (dx/dtheta)

slope_expr = dy_dtheta / dx_dtheta

Evaluate at theta = pi/4

theta_val = sp.pi / 4 slope_val = slope_expr.subs(theta, theta_val) print(f"Slope value: {slope_val}") slope_simplified = sp.simplify(slope_val) print(f"Simplified slope: {slope_simplified}")

[/PYTHON] [PYTHON] import sympy as sp

theta = sp.symbols('theta') r = 2 + sp.cos(theta)

Area

a = sp.integrate((1/2) * r**2, (theta, 0, 2*sp.pi)) print(f"a = {a}")

Slope

x = r * sp.cos(theta) y = r * sp.sin(theta) dx = sp.diff(x, theta) dy = sp.diff(y, theta) slope = dy/dx slope_at_pi_4 = slope.subs(theta, sp.pi/4) simplified_slope = sp.simplify(slope_at_pi_4) print(f"Slope at pi/4 = {simplified_slope}")

Extract b and c

We expect form b + sqrt(c)

Let's try to match it

simplified_slope should be -2 + sqrt(2)

b = -2 c = 2 val = b + sp.sqrt(c) check = sp.simplify(simplified_slope - val) print(f"Check if slope is -2 + sqrt(2): {check == 0}")

Final calculation

result = (a * b * c) / sp.pi print(f"Result = {result}") [/PYTHON] [PYTHON] import sympy as sp

Define symbols

theta = sp.symbols('theta') r = 2 + sp.cos(theta)

Part 1: Area

Area = 1/2 * integral(r^2 dtheta) from 0 to 2pi

area = sp.integrate((1/2) * r**2, (theta, 0, 2*sp.pi)) print(f"Area a = {area}")

Part 2: Slope

x = r cos(theta), y = r sin(theta)

x = r * sp.cos(theta) y = r * sp.sin(theta)

dy/dx = (dy/dtheta) / (dx/dtheta)

dy_dtheta = sp.diff(y, theta) dx_dtheta = sp.diff(x, theta) slope_expr = dy_dtheta / dx_dtheta

Evaluate at theta = pi/4

slope_val = slope_expr.subs(theta, sp.pi/4) slope_simplified = sp.simplify(slope_val) print(f"Slope at pi/4 = {slope_simplified}")

Check form b + sqrt(c)

We suspect b = -2, c = 2

b = -2 c = 2 target = b + sp.sqrt(c) if sp.simplify(slope_simplified - target) == 0: print(f"Confirmed: Slope is {b} + sqrt({c})") else: print("Mismatch in slope form")

Compute final answer

final_val = (area * b * c) / sp.pi print(f"Final value abc/pi = {final_val}") [/PYTHON] The curve $C$ is given by $r = 2 + \cos\theta$ for $0 \le \theta \le 2\pi$.

(1) Area of the enclosed region: The area $a$ is given by the integral: [ a = \frac{1}{2} \int_{0}^{2\pi} r^2 , d\theta = \frac{1}{2} \int_{0}^{2\pi} (2 + \cos\theta)^2 , d\theta ] Expanding the integrand: [ (2 + \cos\theta)^2 = 4 + 4\cos\theta + \cos^2\theta ] Integrating term by term: [ \int_{0}^{2\pi} 4 , d\theta = 8\pi ] [ \int_{0}^{2\pi} 4\cos\theta , d\theta = 0 ] [ \int_{0}^{2\pi} \cos^2\theta , d\theta = \int_{0}^{2\pi} \frac{1 + \cos(2\theta)}{2} , d\theta = \left[ \frac{\theta}{2} + \frac{\sin(2\theta)}{4} \right]_{0}^{2\pi} = \pi ] Thus, the integral is $8\pi + 0 + \pi = 9\pi$. The area is: [ a = \frac{1}{2}(9\pi) = \frac{9\pi}{2} ]

(2) Slope of the tangent line: We need the slope at $(r, \theta) = \left(\frac{4+\sqrt{2}}{2}, \frac{\pi}{4}\right)$. The parametric equations are: [ x = r\cos\theta = (2+\cos\theta)\cos\theta = 2\cos\theta + \cos^2\theta ] [ y = r\sin\theta = (2+\cos\theta)\sin\theta = 2\sin\theta + \sin\theta\cos\theta = 2\sin\theta + \frac{1}{2}\sin(2\theta) ] Differentiating with respect to $\theta$: [ \frac{dx}{d\theta} = -2\sin\theta - 2\cos\theta\sin\theta = -2\sin\theta - \sin(2\theta) ] [ \frac{dy}{d\theta} = 2\cos\theta + \cos(2\theta) ] At $\theta = \frac{\pi}{4}$: [ \frac{dx}{d\theta} = -2\left(\frac{\sqrt{2}}{2}\right) - \sin\left(\frac{\pi}{2}\right) = -\sqrt{2} - 1 ] [ \frac{dy}{d\theta} = 2\left(\frac{\sqrt{2}}{2}\right) + \cos\left(\frac{\pi}{2}\right) = \sqrt{2} + 0 = \sqrt{2} ] The slope $m$ is: [ m = \frac{dy/d\theta}{dx/d\theta} = \frac{\sqrt{2}}{-\sqrt{2}-1} = \frac{\sqrt{2}}{-(1+\sqrt{2})} ] Rationalizing the denominator: [ m = \frac{\sqrt{2}(\sqrt{2}-1)}{-(1+\sqrt{2})(\sqrt{2}-1)} = \frac{2-\sqrt{2}}{-(2-1)} = \frac{2-\sqrt{2}}{-1} = -2 + \sqrt{2} ] This is in the form $b + \sqrt{c}$ with integers $b = -2$ and $c = 2$.

Final Calculation: We have $a = \frac{9\pi}{2}$, $b = -2$, and $c = 2$. We need to compute $\frac{abc}{\pi}$: [ \frac{abc}{\pi} = \frac{\left(\frac{9\pi}{2}\right)(-2)(2)}{\pi} = \frac{-18\pi}{\pi} = -18 ]

[THOUGHT] The calculations are verified. Area $a = 9\pi/2$. Slope $m = -2 + \sqrt{2}$. $b = -2, c = 2$. Expression $\frac{abc}{\pi} = -18$. [/THOUGHT]

\boxed{-18}

\boxed{-18}